~~~~~~~~~~
$T=\frac{4}{1-a}-\frac{1}{a}+\frac{4}{1-b}-\frac{1}{b}+\frac{4}{1-c}-\frac{1}{c}$
$=\frac{5a-1}{a-a^2}+\frac{5b-1}{b-b^2}+\frac{5c-1}{c-c^2}$
Dự đoán dấu = xảy ra khi $a=b=c=\frac{1}{3}$
$\frac{5a-1}{a-a^2} \le ma+n$
<=> $5a-1 \le ma^2 +na-ma^3-na^2$
<=> $ma^3+a^2(n-m)+a(5-n)-1 \le 0$
Xét $f(a)=ma^3+a^2(n-m)+a(5-n)-1$
$f'(a)=3a^2m+2a(n-m+5-n$
Dấu = xảy ra khi $a=\frac{1}{3}$
=> $\begin{cases}f(\frac{1}{3})=0 \\ f'(\frac{1}{3})=0 \end{cases}$
=> $\begin{cases}\frac{2}{27}m+\frac{2}{9}n=\frac{2}{3} \\ \frac{1}{3}m+\frac{1}{3}n=5 \end{cases}$
=> $\begin{cases}m=18 \\ n=-3 \end{cases}$
~~~~~~~~~~
Ta chứng minh: $\frac{5a-1}{a-a^2}\le 18a-3$